Question

Help with integrating

Answer 1

\[\int\limits_{}^{} \frac{ 1 }{ x \sqrt{4x^2 + 9} } dx\]

Answer 2

have you tried a substitution yet?

Answer 3

this thing looks secanty to me for some reason

Answer 4

i tried u substitution and didn't get very far. so i should use trig?

Answer 5

i think a trig might be promising yes

Answer 6

so i could do (4/9)x^2 = tan^2u?

Answer 7

let me see if i can recall the derivative im thinking of:
\[y=sec^{-1}(x)\]\[sec(y)=x\]
\[sec(y)tan(y)y'=1\]
\[y'=\frac{1}{sec(y)tan(y)}\]
\[y'=\frac{1}{x\sqrt{x^2+1}}\]

Answer 8

.... x^2 - 1 that is

Answer 9

that's interesting I haven't seen someone do it like that before. so we are looking for y?

Answer 10

would it then be sec^-1(2x) + c?

Answer 11

i'm a little confused :S

Answer 12

we are looking for a function that will derive down to this basic set up; sec^-1 was what i thought at first, but its not quite right :)

Answer 13

your good with the tan^2 substitution

Answer 14

\[x=\frac32tan(u)\]
\[dx=\frac32sec^2(u)~du\]

Answer 15

ok that makes sense

Answer 16

\[\int\frac{\frac32sec^2(u)}{ \frac32tan(u) \sqrt{9tan^2(u) + 9} } du\]
\[\int\frac{\frac32sec^2(u)}{ \frac32tan(u) *3\sqrt{tan^2(u) + 1} } du\]
\[\int\frac{\frac32sec^2(u)}{ \frac32tan(u) *3sec(u) } du\]
\[\int\frac{\frac32sec(u)}{ \frac92tan(u)} du\]
\[\int\frac{sec(u)}{3tan(u)} du\]

Answer 17

i am now at \[\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ \sec u }{ \tan u } du\]

Answer 18

now into sin and cos?

Answer 19

\[\frac{\frac1{cos}}{\frac{sin}{cos}}=csc\]

Answer 20

so far so good :)

Answer 21

i don't know the integral of csc :/

Answer 22

its the same trick as applied to the integration of sec

Answer 23

oh ok

Answer 24

but what do I do instead of sec x + tan x / sec x + tan x ?

Answer 25

multiply by (csc+cot), prolly got some signs wrong on that tho

Answer 26

csc^2+csc cot ints to -> -cot - csc

so we need to introduce a negative, attach it to the 1/3 outside

Answer 27

\[D[cot+csc]=-csc^2-csc~cot=-(csc^2+csc~cot)\]

Answer 28

you recall what to do next?

Answer 29

i do the sec trick differently

Answer 30

yes\[k\int csc(u)=-k\int\frac{csc^2(u)+csc(u)cot(u)}{csc(u)+cot(u)}\to\ -k\int \frac{v'}{v}\]

Answer 31

oh i see

Answer 32

so you just multiply by (csc u + cot u) / (csc u + cot u)

Answer 33

yep, with a slight adjustment. when we play with trig function that begin with a "c" in them, we introduce a negative into the mix

Answer 34

why?

Answer 35

its just a pattern from derivative\[cos'=-sin\]\[cot'=-csc^2\]\[csc'=-csc~cot\]

Answer 36

oh ok i see. why do we take out the negative if we can easily see that anti d of
-csc^2 is cot?

Answer 37

.... just tripping over my fingers :)

we need to modify the setup as this:\[k\int csc(u)\]
\[k\int\frac{(-1)*(-1)*(csc^2(u)+csc(u)cot(u))}{csc(u)+cot(u)}\]
\[k*(-1)\int\frac{-csc^2(u)-csc(u)cot(u))}{csc(u)+cot(u)}\]
\[-k\int \frac{v'}{v}\]

thats better

Answer 38

haha oh my!

Answer 39

so can we substitute some variable as csc u + cot u

Answer 40

oh that's v

Answer 41

we can if we want, but i just rewrote it in terms of functions hoping that youd see it solution as a pattern

Answer 42

so - 1/3 (ln(cot u + csc u)) + c?

Answer 43

correct, then all thats left is to rework the trigs into their x counterparts

Answer 44

ahhh :) thank you!

Answer 45

and i do see the pattern, it's basically the opposite of the sec trick

Answer 46

yeah, and the first time i seened that sec trick i could have screamed !! lol

Answer 47

so eventually this will come naturally to me? haha because I am just learning trig substitution and it's taking me quite a while to do some of these problems

Answer 48

with alot of practice, itll become somewhat natural yes

Answer 49

so there was no simpler way of doing this problem?

Answer 50

not that i am aware of. im sure the table method (looking it up in a table of derivatives) would have been quicker.

Answer 51

thanks for explaining everything thoroughly instead of just giving the answer

Answer 52

youre welcome :) good luck